SOLUTION: A total of $9000 $⁢9000 is invested: part at 6% 6% and the remainder at 11% 11% . How much is invested at each rate if the annual interest is $600 $⁢600 ?

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Question 1146729: A total of
$9000
$⁢9000
is invested: part at
6%
6%
and the remainder at
11%
11%
. How much is invested at each rate if the annual interest is
$600
$⁢600
?

Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


(1) A setup for a traditional algebraic solution....

x amount at 6%, plus ($9000-x) at 11%, yields $600 interest:

.06%28x%29%2B.11%289000-x%29+=+600

Solving involves only basic algebra; but the decimals make the calculations a bit ugly. I leave it to you to finish the solution by that path.

Here is a faster way to the answer for a "mixture" problem like this.

$9000 all at 6% --> $540 interest
$9000 mixed --> $600 interest
$9000 all at 11% --> $990 interest

Look where the actual interest of $600 lies between $540 and $990:

540 to 990 = 450; 540 to 600 = 60; 60/450 = 2/15.

$600 is 2/15 of the way from $540 to $990; that means 2/15 of the total is invested at the higher rate.

ANSWER: 2/15 of the $9000, or $1200, at 11%; the other $7800 at 6%.

CHECK: .06(7800)+.11(1200) = 468+132 = 600